# asymptotic growth of zeta function on the real line

let $r$ be real $r> 1$ then $$\zeta(r) = O\left(1+\frac{1}{r-1}\right)$$

Can you tell me how to prove this formula?

I found this after posting $$\sum_{n \ge 1} \frac{1}{n^r} = 1 + \sum_{n \ge 2} \frac{1}{n^r} \le 1 + \int_2^\infty \frac{dx}{x^r} = 1 - \frac{1}{r-1} \frac{1}{2}$$ and it seems a right idea but the sign is wrong.

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$1 - \frac{1}{r-1} \frac{1}{2} \leq 1 + \frac{1}{r-1}$ –  Antonio Vargas Feb 4 '13 at 15:47

## 1 Answer

Sure, it comes from Laurent's formula $$\zeta(r) = \frac{r}{r-1} - r\int_1^\infty \frac{\{x\}}{x^{1+r}}dx$$

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thanks, I've got that formula. but how? –  user58512 Feb 4 '13 at 15:46